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Let $(\mathcal{M},g)$ be a compact and connected Riemannian manifold, $\mathrm{d}$ and $\delta$ differential and codifferential, respectively, and $\Delta:=\delta\mathrm{d}+\mathrm{d}\delta$ the corresponding de Rham-Hodge Laplacian. Is it true that

$$\mathrm{ker}(\delta)=\mathrm{ran}(\delta)\oplus\mathrm{ker}(\Delta)$$

Direction "$\supset$" is clear, since $\delta^{2}=0$ and every harmonic form is both closed and coclosed. However, I am struggeling with the other direction. Let $\omega\in\mathrm{ker}(\delta)$. My idea was to use condradiction: Assume $\omega\notin\mathrm{ran}(\delta)\oplus\mathrm{ker}(\Delta)$. Is it possible to argue by Hodge decomposition that then $\omega\in\mathrm{ran}(\mathrm{d})$? Because then the claim follows, since $\omega$ would be both closed and coclosed and hence in particular harmonic.

EDIT: At least for $1$-forms, it should be true: Take $A\in\Omega^{1}(\mathcal{M})$ such that $\delta A=0$. By Hodge decomposition, $A=\mathrm{d}f+\delta F+h$ for $h$ harmonic. Then $\delta A=\Delta f=0$ and hence $f=\mathrm{const}$, since any harmonic $1$-form on a compact manifold is necessarily constant. It follows that $A=\delta F+h\in\mathrm{ran}(\delta)\oplus\mathrm{ker}(\Delta)$.

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2 Answers 2

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You can use a similar argument as in your EDIT: Take $\omega\in\Omega^{k}(\mathcal{M})$ such that $\delta\omega=0$. By Hodge decomposition,

$$\omega=\mathrm{d}\alpha+\delta\beta+h$$

for $\alpha\in\Omega^{k-1}(\mathcal{M})$, $\beta\in\Omega^{k+1}(\mathcal{M})$ and $h\in\mathcal{H}^{k}(\mathcal{M})$. It follows that

$$\delta\omega=\delta\mathrm{d}\alpha\stackrel{!}{=}0$$

$\delta\mathrm{d}$ is for $k\neq 0$ different from $\Delta$. However, since also $\mathrm{d}^{2}\alpha=0$, you see that $\mathrm{d}\alpha$ is both closed and coclosed and hence in particular harmonic, i.e. $\mathrm{d}\alpha\in\mathcal{H}^{k}(\mathcal{M})$. This shows that

$$\omega=\delta\beta+(h+\mathrm{d}\alpha)$$

gives the required decomposition.

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Since $\delta$ is the formal adjoint of $d$, for any form $\alpha$ satisfying $\delta d \alpha = 0$ we have $$\langle d\alpha, d\alpha \rangle = \langle \delta d \alpha, \alpha\rangle = 0,$$ which implies $$d \alpha = 0.$$ So, your argument for $1$-forms holds for a general form.

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